AcWing 903. Expensive bride price solution (the shortest path - building map, Dijkstra)

AcWing 903. Expensive betrothal gifts
The difficulty of this problem is building a map , The trick is to establish a virtual starting point , Let each road become a virtual starting point （ In the picture S, Points in the code 0） Reach the end point （ spot 1） A path for , Then find the shortest path

#include<bits/stdc++.h>

using namespace std;

const int N = 110, INF = 0x3f3f3f3f;

#define rep(i, a, b) for(int i = a; i <= b; i ++ )
#define rop(i, a, b) for(int i = a; i < b; i ++ )

int w[N][N];
int level[N];  // Master level 
int d[N];
int n, m;
bool st[N];

int dijkstra(int down, int up){
    
	
	memset(d, 0x3f, sizeof d);
	memset(st, 0, sizeof st);
	
	d[0] = 0;
	rep(i, 1, n){
    
		int t = -1;
		rep(j, 0, n){
      // What we are looking for here is Li i Point the nearest point , So include the virtual starting point 0 
			if(!st[j] && (t == -1 || d[t] > d[j])) t = j;  // This step is to find the point with the smallest distance at present and update the distance of other points , So we must first judge whether this point has been used  
		} 
		st[t] = true;  // Marking this point has been used , Because each point is the closest when it is first used , Later, it will not be updated to a smaller distance  
		rep(j, 1, n){
    
			if(level[j] >= down && level[j] <= up){
    
				d[j] = min(d[j], d[t] + w[t][j]);
			}
		}
	}
	return d[1];
}

int main()
{
    
	
	cin>>m>>n;
	memset(w, 0x3f, sizeof w);
	rep(i, 1, n) w[i][i] = 0;
	rep(i, 1, n){
    
		int price, cnt;
		cin>>price>>level[i]>>cnt;
		w[0][i] = min(w[0][i], price);
		while(cnt -- ){
    
			int pr, id;
			cin>>id>>pr;
			w[id][i] = min(w[id][i], pr);  // The question is id Can replace i, When building the map id towards i The edge of , So it should be w[id][i]
		}
	}
	
	int res = INF;
	rep(i, level[1] - m, level[1]){
    
		res = min(res, dijkstra(i, i + m));
	}
	
	cout<<res<<endl;
	
	return 0;
}